03_sorting and it's types with example .ppt

Sorting Algorithms
CENG 213
METU/ODTÜ
Data Structures
Prof. Dr. Yusuf Sahillioğlu

Sorting
• Sorting is a process that organizes a collection of data
into either ascending or descending order.
• Formally
• Input: A sequence of n numbers <a1,a2,…,an>
• Output: A reordering <a’1,a’2,…,a’n> of the sequence such
that a’1 ≤ a’2 ≤ … ≤ a’n
• Given the input <6, 3, 1, 7>, the algorithm should produce
<1, 3, 6, 7>
• Called an instance of the problem

Sorting
• Sorting is a process that organizes a collection of data
into either ascending or descending order.
• We encounter sorting almost everywhere:
– Sorting prices from lowest to highest
– Sorting flights from earliest to latest
– Sorting grades from highest to lowest
– Sorting songs based on artist, album, playlist,
etc.

Sorting Algorithms
• There are many sorting algorithms (as of
27.10.2014 there are 44 Wikipedia entries)
• In this class we will learn:
– Selection Sort
– Insertion Sort
– Bubble Sort
– Merge Sort
– Quick Sort
– Bogo sort and Sleep sort are some bad/slow algorithms
• These are among the most fundamental sorting
algorithms

Sorting Algorithms
• As we learnt in the Analysis lecture (time
complexity), a stupid approach uses up
computing power faster than you might think.
• Sorting a million numbers:
• Interactive graphics: Algorithms must
terminate in 1/30 of a sec.

Sorting Algorithms
• As we learnt in the Analysis lecture (time
complexity), a stupid approach uses up
computing power faster than you might think.
• Sorting a million numbers:
• https://youtu.be/kPRA0W1kECg

Selection Sort
• Partition the input list into a sorted and unsorted
part (initially sorted part is empty)
• Select the smallest element and put it to the end of
the sorted part
• Increase the size of the sorted part by one
• Repeat this n-1 times to sort a list of n elements

23 78 45 8 32 56
8 78 45 23 32 56
8 23 45 78 32 56
8 23 32 78 45 56
8 23 32 45 78 56
8 23 32 45 56 78
Original List
After pass 1
After pass 2
After pass 3
After pass 4
After pass 5
Sorted Unsorted

Selection Sort (cont.)
template <class Item>
void selectionSort(Item a[], int n) {
for (int i = 0; i < n-1; i++) {
int min = i;
for (int j = i+1; j < n; j++)
if (a[j] < a[min]) min = j;
swap(a[i], a[min]);
}
}
template < class Object>
void swap( Object &lhs, Object &rhs )
{
Object tmp = lhs;
lhs = rhs;
rhs = tmp;
}

Selection Sort -- Analysis
• What is the complexity of selection sort?
• Does it have different best, average, and worst case
complexities?

Selection Sort – Analysis (cont.)
• Selection sort is O(n2
) for all three cases (prove this)
• Therefore it is not very efficient

Insertion Sort
• Insertion sort is a simple sorting algorithm that is
appropriate for small inputs.
– Most common sorting technique used by card players.
• Again, the list is divided into two parts: sorted and
unsorted.
• In each pass, the first element of the unsorted part
is picked up, transferred to the sorted sublist, and
inserted at the appropriate place.
• A list of n elements will take at most n-1 passes to
sort the data.

Original List
After pass 1
After pass 2
After pass 3
After pass 4
After pass 5
23 78 45 8 32 56
23 78 45 8 32 56
23 45 78 8 32 56
8 23 45 78 32 56
8 23 32 45 78 56
8 23 32 45 56 78
Sorted Unsorted

Insertion Sort Algorithm
void insertionSort(Item a[], int n)
{
for (int i = 1; i < n; i++)
{
Item tmp = a[i]; // the element to be inserted
int j;
for (j=i; j>0 && tmp < a[j-1]; j--)
a[j] = a[j-1]; // shift elements
a[j] = tmp; // insert
}
}

Insertion Sort – Analysis
• Running time depends on not only the size of the array but also
the contents of the array.
• Best-case:  O(n)
– Array is already sorted in ascending order.
• Worst-case:  O(n2
)
– Array is in reverse order:
• Average-case:  O(n2
)
– We have to look at all possible initial data organizations.

Analysis of insertion sort
• Which running time will be used to characterize this
algorithm?
– Best, worst or average?
• Worst:
– Longest running time (this is the upper limit for the algorithm)
– It is guaranteed that the algorithm will not be worse than this.
• Sometimes we are interested in average case. But there are
some problems with the average case.
– It is difficult to figure out the average case. i.e. what is average
input?
– Are we going to assume all possible inputs are equally likely?
– In fact for most algorithms average case is same as the worst case.

Bubble Sort
• Repeatedly swap adjacent elements that are out of
order.
• https://youtu.be/vxENKlcs2Tw
• Sort pairs of elements far apart from each other, then
progressively reduce the gap between elements to be
compared. Starting with far apart elements, it can
move some out-of-place elements into position faster
than a simple nearest neighbor exchange. This
generalization is called Shell Sort.

Bubble Sort
23 78 45 8 32 56
23 45 78 8 32 56
23 45 8 78 32 56
23 45 8 32 78 56
23 45 8 32 56 78
23 8 45 32 56 78
8 23 45 32 56 78
8 23 32 45 56 78

Bubble Sort Algorithm
void bubleSort(Item a[], int n)
{
bool sorted = false;
int last = n-1;
for (int i = 0; (i < last) && !sorted; i++){
sorted = true;
for (int j=last; j > i; j--)
if (a[j-1] > a[j]{
swap(a[j],a[j-1]);
sorted = false; // signal exchange
}
}
}

Bubble Sort – Analysis
• Best-case:  O(n)
– Array is already sorted in ascending order.
• Worst-case:  O(n2
)
– Array is in reverse order:
• Average-case:  O(n2
)
– We have to look at all possible initial data organizations.
• In fact, any sorting algorithm which sorts elements by swapping
adjacent elements can be proved to have an O(n2
) average case
complexity

Theorem
• Any sorting algorithm which sorts elements by swapping
adjacent elements can be proved to have an O(n2
) average case
complexity
• To understand this, consider the following array:
• [3, 2, 1]
• Here, we have three inversions: (3, 2), (3, 1), (2, 1)
• An inversion is defined as a pair (a, b) where a > b and index(a) <
index(b)
• Note that a single swap of adjacent elements can only remove
one inversion. Thus, for the above example we need 3 swaps to
make the array sorted

Theorem
• Generalizing this, for an array of n elements, there can be C(n,2)
total pairs (where C represents combination)
• C(n,2) = n(n-1) / 2
• We can assume that, on average, half of these pairs are
inversions.
• Average number of inversions = n(n-1) / 4
• As each swap of adjacent elements can remove a single
inversion, we need n(n-1) / 4 swaps to remove all inversions (that
is to make the array sorted)
• The complexity of n(n-1) / 4 swaps is equal to O(n2
)
• This is the lower bound on the average case complexity of sorting
algorithms that sort by swapping adjacent elements

Recursive Insertion Sort
• To sort A[1..n], we recursively sort A[1..n-1] and then insert
A[n] into the sorted array A[1..n-1]
• A good practice of recursion
• Code?

• To sort A[1..n], we recursively sort A[1..n-1] and then insert A[n]
into the sorted array A[1..n-1]
insertionSortRec(A, n)
{
if (n>1)
insertionSortRec(A, n-1)
insertKeyIntoSubarray(A[n], A[1..n-1]) //trivial; can
//be done
in //O(n)
}

• A = 5 2 4 6 1 3
insertKeyIntoSubarray
1 2 4 5 6 3
2 4 5 6 1
2 4 5 6
2 5 4
5 2

• T(n) = T(n-1) + O(n) is the time complexity of this sorting

• T(n) = T(n-1) + O(n)
• Reduce the problem size to n-1 w/ O(n) extra work
• Seems to be doing O(n) work n-1 times; so O(n2
) guess?
• T(n) ≤ cn2
//assume holds
T(n) ≤ c(n-1)2
+ dn
= cn2
-2cn + c + dn
= cn2
–c(2n -1) + dn
≤ cn2
//because large values of c dominates d

Recurrences
• How about the complexity of T(n) = T(n/2) + O(1)
• Reduce the problem size to half w/ O(1) extra work
• Seems to be doing O(1) work logn times; so O(logn) guess?
• T(n) ≤ clogn//assume holds
T(n) ≤ c(log(n/2)) + d
= clogn– clog2 + d
= clogn– e //c can always selected to be > constant d
≤ clogn

Recurrences
• How about the complexity of T(n) = 2T(n/2) + O(n)
• Reduce the problem to 2 half-sized problems w/ n extra work
• Seems to be doing O(n) work logn times; so O(nlogn) guess?
• T(n) ≤ cnlogn//assume holds
T(n) ≤ 2c(n/2 log(n/2)) + dn
= cnlogn– cnlog2 + dn
= cnlogn– n(c’ + d) //c’ = clog2
≤ cnlogn

Recurrences
• How about the complexity of T(n) = T(n-1) + T(n-2)
• Reduce? the problem to a twice bigger one w/ no extra work
• T(n-1) + T(n-2) > 2T(n-2)  n replaced by 2n-4 (doubled)
• Or, n-size problem replaced with 2n-3 size problem (doubled)
• Seems to be doubling the problem size; so O(2n
) is good guess

Fast Algorithm
• Describe an algorithm that, given a set S of n integers and
another integer x, determines whether or not there exists 2
elements in S whose sum is exactly x
• O(n2
) //brute force

Fast Algorithm
• Describe an algorithm that, given a set S of n integers and
another integer x, determines whether or not there exists 2
elements in S whose sum is exactly x
• O(nlogn) //sort followed by binary search

Mergesort
• Mergesort algorithm is one of two important divide-and-conquer
sorting algorithms (the other one is quicksort).
• It is a recursive algorithm.
– Divides the list into halves,
– Sort each halve separately (recursively), and
– Then merge the sorted halves into one sorted array.
• https://youtu.be/es2T6KY45cA

Mergesort - Example
6 3 9 1 5 4 7 2
5 4 7 2
6 3 9 1
6 3 9 1 7 2
5 4
6 3 1
9 5 4 2
7
3 6 1 9 2 7
4 5
2 4 5 7
1 3 6 9
1 2 3 4 5 6 7 9
divide
divide
divide
divide
divide
divide
divide
merge merge
merge
merge
merge merge
merge

Merge
const int MAX_SIZE = maximum-number-of-items-in-array;
void merge(DataType theArray[], int first, int mid, int last)
{
DataType tempArray[MAX_SIZE]; // temporary array
int first1 = first; // beginning of first subarray
int last1 = mid; // end of first subarray
int first2 = mid + 1; // beginning of second subarray
int last2 = last; // end of second subarray
int index = first1; // next available location in tempArray
for ( ; (first1 <= last1) && (first2 <= last2); ++index)
if (theArray[first1] < theArray[first2])
tempArray[index] = theArray[first1++];
else
tempArray[index] = theArray[first2++];

Merge (cont.)
// finish off the first subarray, if necessary
for (; first1 <= last1; ++first1, ++index)
tempArray[index] = theArray[first1];
// finish off the second subarray, if necessary
for (; first2 <= last2; ++first2, ++index)
tempArray[index] = theArray[first2];
// copy the result back into the original array
for (index = first; index <= last; ++index)
theArray[index] = tempArray[index];
}

Mergesort
void mergesort(DataType theArray[], int first, int last) {
if (first < last) { // more than one item
int mid = (first + last)/2; // index of midpoint
mergesort(theArray, first, mid);
mergesort(theArray, mid+1, last);
// merge the two halves
merge(theArray, first, mid, last);
}
} // end mergesort

Analysis of Mergesort
• What is the complexity of the merge operation for merging two
lists of size n/2?
• It is O(n) as we need to copy all elements
• Then the complexity of mergesort can be defined using the
following recurrence relation:
T(n) = 2T(n/2) + n
• Solving this relation gives us O(nlogn) complexity //Slide 30
• The complexity is the same for the best, worst, and average
cases
• The disadvantage of mergesort is that we need to use an extra
array for the merge operation (not memory efficient)

Analysis of Mergesort
• Solving T(n) = 2T(n/2) + n gives us O(nlogn) complexity

Merging Step (Linear time)
• Extra array for merge operation

Demo
• On array = {5, 2, 4, 7, 1, 3, 2, 6}

Quicksort
• Like mergesort, quicksort is also based on
the divide-and-conquer paradigm.
• But it uses this technique in a somewhat opposite manner,
as all the hard work is done before the recursive calls.
• It works as follows:
1. First selects a pivot element,
2. Then it partitions an array into two parts (elements smaller than and
greater than or equal to the pivot)
3. Then, it sorts the parts independently (recursively),
4. Finally, it combines the sorted subsequences by
a simple concatenation.
• https://youtu.be/vxENKlcs2Tw

Partition
• Partitioning places the pivot in its correct place position within the array.
• Arranging the array elements around the pivot p generates two smaller sorting
problems.
– sort the left section of the array, and sort the right section of the array.
– when these two smaller sorting problems are solved recursively, our bigger
sorting problem is solved.

Pivot Selection
• Which array item should be selected as pivot?
– Somehow we have to select a pivot, and we hope that we will
get a good partitioning.
– If the items in the array arranged randomly, we choose a pivot
randomly.
– We can choose the first or last element as the pivot (it may not
give a good partitioning).
– We can choose the middle element as the pivot
– We can use a combination of the above to select the pivot (in
each recursive call a different technique can be used)

Partitioning
Initial state of the array

Partitioning
Moving theArray[firstUnknown] into S1 by swapping it with
theArray[lastS1+1] and by incrementing both lastS1 and firstUnknown.

Partitioning
Moving theArray[firstUnknown] into S2 by incrementing firstUnknown.

Partition Function
template <class DataType>
void partition(DataType theArray[], int first, int last,
int &pivotIndex) {
int pIndex = choosePivot(theArray, first, last);
// put pivot at position first
swap(theArray[pIndex], theArray[first]);
DataType pivot = theArray[first]; // copy pivot

Partition Function (cont.)
int lastS1 = first; // index of last item in S1
int firstUnknown = first + 1; //index of 1st item in unknown
// move one item at a time until unknown region is empty
for (; firstUnknown <= last; ++firstUnknown) {
if (theArray[firstUnknown] < pivot)
++lastS1;
swap(theArray[firstUnknown], theArray[lastS1]);
}
// place pivot in proper position and mark its location
swap(theArray[first], theArray[lastS1]);
pivotIndex = lastS1;
}

Quicksort Function
void quicksort(DataType theArray[], int first, int last) {
int pivotIndex;
if (first < last) {
partition(theArray, first, last, pivotIndex);
// sort regions S1 and S2
quicksort(theArray, first, pivotIndex-1);
quicksort(theArray, pivotIndex+1, last);
}
}

Quicksort – Analysis
• If we always select the smallest or largest element as the pivot,
we’ll not be able to divide the array into similar sized partitions
• In that case, the complexity of the quicksort can be defined by:
T(n) = n + T(1) + T(n-1)
• This gives O(n2
) complexity (worst case) //Slide 28
• If our partitions are equal sized we have:
T(n) = n + 2T(n/2) (same as mergesort)
• This gives O(nlogn) complexity (best case)
• On average, quicksort has been proven to have O(nlogn)
complexity as well (based on alternation b/w good/bad partitions)
• It also does not need an extra array like mergesort
• Therefore, it is the most popular sorting algorithm

C/C++ Language Support
• We can implement sorting algorithms ourselves
• We can also use existing implementations
• In C, the function qsort (part of stdlib.h header) implements the
quicksort algorithm
• In C++ std::sort (part of algorithm header) implements a mixture
of quicksort, heapsort, and insertion sort
• Many sorting algorithms in comparison:
https://youtu.be/ZZuD6iUe3Pc

Sorting vs. Geometry
• A cool reduction of Convex Hull problem into Sorting problem
• This is the convex hull of the point set P:
• Release a large rubber band around points/nails
• Given unsorted a0, a1, a2, .., an, construct set of points in the
plane via ai=ai2
parabola as shown at left
• Every point must be on the convex hull ‘cos parabolas are convex
• Convex hull of this parabola = Ordered hull vertices = Sorting

03_sorting and it's types with example .ppt

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